Question

Giải pt:

Answer 1

đk |x| khác 2

ta có

: \(46\left(\dfrac{x^2-9}{x^2-4}\right)=46\left[\dfrac{x-3}{x+2}.\dfrac{x+3}{x-2}\right]\)

đặt \(a=\dfrac{x-3}{x+2};b=\dfrac{x+3}{x-2}\)

<=> 3b^2 +168a^2-46ab=0

<=>\(\left(\sqrt{3}b-\dfrac{23\sqrt{3}}{3}a\right)^2+\left(168-\dfrac{23^2}{3}\right)a^2=0\)

\(\left(\sqrt{3}b-\dfrac{23\sqrt{3}}{3}a-\dfrac{5\sqrt{3}}{3}a\right)\left(\sqrt{3}b-\dfrac{23\sqrt{3}}{3}a+\dfrac{5\sqrt{3}}{3}a\right)=0\)\(\begin{matrix}3b-28a=0\\b-6a=0\end{matrix}\)

\(\left\{{}\begin{matrix}\left|x\right|\ne2\\\left[{}\begin{matrix}3\left(x+3\right)\left(x+2\right)=28\left(x-3\right)\left(x-2\right)\\\left(x+3\right)\left(x+2\right)=6\left(x-3\right)\left(x-2\right)\end{matrix}\right.\end{matrix}\right.\)

tự làm tiếp